I came across this while doing questions on field axioms. I need this proved for where $m,n \in \mathbb Z$ and $a \neq 0$. Please do it in complete steps. I looked this up on the site but I couldn't understand what's been done.
How to prove $a^ma^n = a^{m+n}$ using field axioms?
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$\begingroup$
real-analysis
algebra-precalculus
exponentiation
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0Is this about elementary exponents ? – 2017-02-06
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0$(a\times a\times....)\times(a\times a\times....)=a\times a\times....a\times a\times....$ by the associative property of the real numbers – 2017-02-06
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0For positive integers $m,n$ this can be shown by (as @tired comments) induction using the associated property. For non-positive integers the field axioms come into additional force in providing that $a\neq 0$ has an inverse, and thus one builds on the positive integer case by observing that $a^{-m}$ is the inverse of $a^m$, etc. – 2017-02-10
3 Answers
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If you are thinking of a real number a and integers m and n it could be easily proven by definition.
So what is a^m: a * a * ... * a m times. (1)
So what is a^n: a * a * ... * a n times. (2)
The same is with a^(m + n): a * a * ... * a m + n times. (3)
When we multiply (1) and (2) we got a * a * ... * a m + n times which is equal to (3).
It means that we proved the identity.
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The definition of $a^m$ is by induction on $m$. So the best way to prove what you want to prove is by induction. As $a^0= 1$by definition, what you want to prove is immediate for $n=0$. The step $n=1$ is interesting for the step $n\to n+1$. $a^{m+1} = a^m a$ by definition and so $a^{m+1} = a^m a^1$. Now the step $n\to n+1$ follows easily from the definitions
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0Can you elaborate more. I'm having trouble understanding. Thank you. – 2017-02-06
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0What part are you having trouble understanding ? (So I know what to elaborate) – 2017-02-06
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Okay I found the answer. You have to consider all the values for m and n such that $m,n\in\mathbb Z^+$, $m,n\in\mathbb Z^-$, $m,n = 0$ and $m\in\mathbb Z^+ and, n\in\mathbb Z^- s.t. m>n$.
